Codeforce - 712 -C. Memory and De-Evolution

C. Memory and De-Evolution
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
Memory is now interested in the de-evolution of objects, specifically triangles. He starts with an equilateral triangle of side length x, and he wishes to perform operations to obtain an equilateral triangle of side length y.

In a single second, he can modify the length of a single side of the current triangle such that it remains a non-degenerate triangle (triangle of positive area). At any moment of time, the length of each side should be integer.

What is the minimum number of seconds required for Memory to obtain the equilateral triangle of side length y?

Input
The first and only line contains two integers x and y (3 ≤ y < x ≤ 100 000) — the starting and ending equilateral triangle side lengths respectively.

Output
Print a single integer — the minimum number of seconds required for Memory to obtain the equilateral triangle of side length y if he starts with the equilateral triangle of side length x.

Examples
input
6 3
output
4
input
8 5
output
3
input
22 4
output
6
Note
In the first sample test, Memory starts with an equilateral triangle of side length 6 and wants one of side length 3. Denote a triangle with sides a, b, and c as (a, b, c). Then, Memory can do .

In the second sample test, Memory can do .

In the third sample test, Memory can do:

.
题意：有一个正三角，边长为 x ，要求一次变换一条边长，并且每一次都满足 non-degenerate triangle ，即满足三角形三边和差定理。使之最终变成边长为 y 的正三角形，求其最少步数。

AC代码思路：这一题用逆向思维更好想，即从 y 变成 x。且每一步的变换最大范围是除去需要变换的边的其他两条边的和 -1 。然后用贪心。

AC代码：

#include <iostream>
using namespace std;

int main()
{
    int x,y;
    cin>>x>>y;
    int cnt=0;
    int a=y,b=y,c=y;
    while (a!=x || b!=x || c!=x)
    {
        if (a!=x) {a=min((b+c-1),x);cnt++;}
        if (b!=x) {b=min((a+c-1),x);cnt++;}
        if (c!=x) {c=min((a+b-1),x);cnt++;}
    }
    cout<<cnt;
}